Prove that if a ring homomorphism is surjective, the associated spectral map is injective

Question

I recently had a bunch of questions on a problem set that I could not solve and the instructor is not providing solutions.

Let $f: A \to B$ be a surjective ring homomorphism. Prove that the induced map $g: \operatorname{Spec} B \to \operatorname{Spec} A$ is injective.

I tried using the definition of injectivity but I don't know where to go from there and how to use the surjectivity of $f$. Can someone give me a detailed explanation?

Answer 1

This is a community wiki post recording the answer from the comments so that this post may be marked as answered.

Assume $f^{-1}(\mathfrak{p})=f^{-1}(\mathfrak{q})$, then we have $\mathfrak{p}=f(f^{-1}(\mathfrak{p}))=f(f^{-1}(\mathfrak{q}))=\mathfrak{q}$ as the map is surjective. – Jiacheng

Answer 2

As a bonus observation: if $f: A \to B$ is injective, then $g$ has a dense image:

Let $D(a) = \{I \in \text{Spec } A\mid a \notin I\}$ be a non-empty basic open subset in the image, for some $a \in A$. Note that $a \neq 0$ as $D(0)=\emptyset$ (all ideals contain $0$).

Then $D(a) \cap g[\text{Spec } B]$ is non-empty iff

$$\exists I' \in \text{Spec } B: g(I')=f^{-1}[I'] \in D(a)$$ or

$$\exists I' \subseteq B \text{ prime ideal }: f(a) \notin I'$$

which is true, as we have $f(a) \neq 0$ by $f$ being an injective ring homomorphism and every non-zero element of a unitary ring has some prime ideal that it is not in. (Well,not quite, I realised it's probably a bit more complicated and I have to consider the radicals of the rings instead but that's ring theory stuff and I'm but a simple topologist, but I did find this answer).